I understand the statement for the mapping property of quotient rings but I want to know the significance of picking an ideal contained in the kernel of R.It seems that this has something to do with the fact there is a further quotient?

For example, our lecturer told us the mapping property is the reason why the following two rings are isomorphic:

  1. R[x]/<2x,3x>

where <2x,3x> denote the ideal generated by the two polynomials 2x and 3x.

  1. (R[x]/<2x>)/<3x> (there seems to be another form for this, which is (R[x]/<2x>)/<3x+<2x>>

I guess my question is why the mapping property has something to do with the fact that it makes no difference to introduce one relation at a time and to introduce two relations together?


Define \begin{align*} \phi : R[x]/(2x)&\to R[x]/(2x,3x)\\ p(x) + (2x)&\mapsto p(x) + (2x,3x). \end{align*} It is not hard to verify that this is a ring homomorphism, and it is clearly surjective. Hence, $(R[x]/(2x))/\ker\phi\cong R[x]/(2x,3x)$ by the first isomorphism theorem.

Any element of the form $3x p(x) + (2x)$ is in the kernel, as $3x p(x)\in (2x,3x)$, so $(3x + (2x))\subseteq\ker\phi$.

Conversely, suppose $p(x) + (2x)\in\ker\phi$. Then $p(x)\in (2x,3x)$ by definition of $R[x]/(2x,3x)$ That is, $p(x) = 2x q(x) + 3x r(x)$ for some polynomials $q$ and $r$ in $R[x]$. But then, $$ p(x) + (2x) = 2x q(x) + 3x r(x) + (2x), $$ and $$ 2x q(x) + 3x r(x) + (2x) = 3x r(x) + (2x), $$ as $2x q(x) + 3x r(x) - 3x r(x) = 2x q(x)\in (2x)$. So $p(x) + (2x) = 3x r(x) + (2x)$, and hence $p(x) + (2x)\in (3x + (2x))$, so $\ker\phi\subseteq (3x + (2x))$.

Then $\ker\phi = (3x + (2x))$, which gives the isomorphism you were wondering about. (As for the "other form" of $(R[x]/(2x))/(3x + (2x))$, it technically doesn't make sense, as $3x$ is not an element of $R[x]/(2x)$, strictly speaking. I would say that $(R[x]/(2x))/(3x)$ is really just shorthand for $(R[x]/(2x))/(3x + (2x))$.)

In general, if you have an ideal $I = (a,b)\subseteq R$, then $R/I\cong (R/(a))/(b + (a))\cong (R/(b))/(a + (b))$. This is akin to saying that if you divide an element by both $a$ and $b$ (maybe pretend $R = \Bbb Z$ for a moment), the remainder is the same if you had divided first by $a$, and then by $b$ or vice versa, and this is also the same as finding the remainder when you divide by their gcd, if it exists.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.